# DM-Sets,Relations and Functions-Q5

+1 vote

Let A and B are 2 sets $|A| = |B|$, there is a one to one function $f$ from A to B.

Which of the following must be true for $f?$

S1: $f$ is onto function

S2: $f$ has an inverse

(A). Only S1

(B). Only S2

(C). Both S1 and S2

(D). None of the above

reshown Jul 6

Above function is one-one but not onto.

answered Jul 9 by (29,920 points)
+1 vote

if $f(\infty)$ = 1 then $f^{-1} (1) = \text{Not defined}$

Hence S2 is false

Also consider a function from $\mathbb{N}\rightarrow \mathbb{N}$

f(i) = 2i

f is one to one but not onto and $|\mathbb{N}| = |\mathbb{N}|$

So S1 is also false

$\text{D is the correct option}$

answered Jul 6 by (970 points)
edited Jul 6
You mean, D is the correct option?
yes
that was a typing error
D is correct one consider both set are infinite N atural number and fun f(I)=2i
answered Jul 6 by (18,160 points)
@ tsnikhilsharmagate2018  D would be a answers as there would be one to one mapping so every elments would be mapped and the cardinality of both the set are same so they must be onto please clarify ?
For i=1 output is 2 for I=2 output is 4 as so on.....
In codomain no element map to 1,3,5 (odd) number it is one to one but not onto
i got it if we select the function like like f(X)=2i then every element would be mapped to one only but some values would be left which would be make as not onto and for that number there would be no inverse
thanks bro .
Yes but if given A and B are finite than both are true.
can't we take function from natural numbers to even numbers or mapping 1 -> 1 or 2->2 for better clarity check this :
https://math.stackexchange.com/questions/1535205/one-to-one-correspondence-between-set-of-positive-rational-numbers-and-set-of-po
hey guys, do you have a group or something like a whatsapp group where you can discuss the questions and other doubts
Rishi we have to find counter examples
yes, you can find... when you found some counterexample in which the size of two sets are equal then there exists one-one and onto
–1 vote
suppose f:A->B such that |A|=m and |B|=n

for one to one function  |B|>=|A|  and for onto function |A|>=|B|

so for bijection |A|=|B|

since it is already given |A|=|B| and one one ,so according to above theory it will be onto . and since it is bijection ,inverse will also exist.

option c
answered Jul 6 by (4,600 points)
D is correct